Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.6 Solving Quadratic Equations by Using the Quadratic Formula - 4.6 Exercises - Page 373: 51

Answer

$f= -10.46, f= 14.46$

Work Step by Step

Use either the square root method, the quadratic formula or factoring. $$ \begin{aligned} 120 & =-28 f+7 f^2-939 \\ 120+939 & =7 f^2-28 f \\ 1059 & =7 f^2-28 f\\ f^2-4 f & =\frac{1059}{7} \\ f^2-4 f+2^2 & =\frac{1059}{7}+2^2 \\ (f-2)^2 & =\frac{1059+4\cdot 7}{7} \\ (f-2)^2 & =\frac{1087}{7} \\ f-2 & = \pm \sqrt{\frac{1087}{7}} \\ f & =2 \pm 12.46. \end{aligned} $$ This gives: $$ \begin{aligned} f & =2-12.46 \\ & =-10.46 \\ f & =2+12.46 \\ & =14.46 \end{aligned} $$
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